Php 从MySQL动态生成HTML表

我有三门课，考试，考试

表格-科目

主体

主题名

受试者

表-检查

埃克萨米德

考试

表格检查

fname

名字

学生

得分

主体

关于subject.subjectExamid=exam.examid

在考试中。主语

现在，我想生成一个HTML表格，显示学生在每一篇论文中针对主题所获得的分数

结构表输出 每个科目分数的学生详细信息

编辑代码示例

<?php
  $examinid = 3;
  $subjects = mysqli_query(
      $con," 
          SELECT * FROM subjects
          WHERE examid = '$examinid'
          ORDER BY shortname ASC

  ");
  $content = mysqli_query(
      $con," 
          SELECT DISTINCT exam.idcandidate, exam.sex, exam.fname, exam.lname
          FROM examinations 
          AS exam
          INNER JOIN examinfo
          AS info
          ON exam.id_subject = info.idsubject
          WHERE info.idexam = '$examinid'         
  ");
?>
<div id="table_1">
  <table cellpadding="0" cellspacing="0" border="0">
      <tr>
          <td class="table1tr">#</td> 
            <td class="table1tr">Candidate</td>   
            <td class="table1tr">ID</td>  
            <td class="table1tr">Sex</td> 
          <?php
              // output subjects 
              while($subRow = mysqli_fetch_array($subjects)){
                  $arbv = strtoupper($subRow['shortname']);
                  $subjectname = ucwords(strtolower($subRow['subjectname']." - ".$subRow['subjectid'].""));
          ?>
          <td class="table1tr" title="<?php echo $subjectname; ?>">
              <?php echo $arbv; ?>
            </td>
          <?php   
              }
          ?>                  
            <td class="table1tr">Exam</td>
        </tr>
      <?php
          while($stdnt = mysqli_fetch_array($content)){
              $fullname = ucwords(strtolower("$stdnt[lname] $stdnt[fname]"));
              $studentid = str_replace(array('/', 'M', 'W', 'S', 'F', '-'), "",$stdnt['idcandidate']);
              if($sex = $stdnt['sex'] == Male){
                  $sex = M;
              }else{ $sex = F; }
              $id_subject = $stdnt['id_subject'];
              $x++;
              $zebra_1 = ($x%2)? 'TableZebra_1': 'TableZebra_2';
      ?>        
      <tr>
          <td class="<?Php echo $zebra_1; ?>"><?php echo $count++; ?></td>    
            <td class="<?Php echo $zebra_1; ?>"><?Php echo $fullname; ?></td> 
            <td class="<?Php echo $zebra_1; ?>"><?php echo $studentid; ?></td>    
            <td class="<?Php echo $zebra_1; ?>"><?php echo $sex; ?></td>
          <td class="<?Php echo $zebra_1; ?>">
            <!-- Problem is here how to output the subject grades $grade -->
            <!-- 
                My first unsuccessful approach  

                SELECT score
                FROM examinations AS test
                INNER JOIN examinfo AS testinfo ON testinfo.idsubject = test.id_subject
                WHERE testinfo.idexam
                IN (

                SELECT idexam
                FROM examinfo
                WHERE idexam = $examinid
                )
                AND test.id_subject = $id_subject AND test.idcandidate = '$studentid'

                Then output results - But this falls it shows one student subjects in one cell
            -->
            </td> 
            <td class="<?Php echo $zebra_1; ?>">Exam</td>
        </tr>        
          <?php   
              } // loop content
          ?>      

    </table>  
 </div>

如果您的解决方案不是concat（）；你可以先

遵循简单的步骤

1 loop $contents // to get info such as studentid
2 inside the loop of $contents loop $subjects // to get all subjects including subjectids
3 inside $subject loop, loop examinations table where studentid = '$studentid' AND subjectid = '$subjectid'
if step three return null echo empty cell otherwise echo cell with score

我没有时间测试这个，但是你可以按照步骤操作，否则它会工作的，试试谷歌搜索
我们不会为你神奇地编写代码。。。如果您想尝试并发布它，那么我很乐意帮助您解决代码中的错误。。但是堆栈溢出不是为了给你的工作编码。。你应该从一些研究开始约翰·鲁德尔检查编辑